So the correct way is to connect Q2 to the 3to8 decoder and that's it?
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SobiaholicNov 11 '12 at 20:50

No, because the 3-to-8 must be active for A3=1, hence for its EN must be OR( Q2, Q3 ), and its third address input must be A2. Why do you want to use such an asymmetric set of chips? Five 2-to-4 decoders would be a more natural choice (or two 3-to-8 and one 1-to-two).
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Wouter van OoijenNov 11 '12 at 21:19

@WoutervanOoijen this is a basic digital logic course. It's not my major, but we have to take it. About your suggestion, Yes! I was thinking inputting A2 is more correct than inputting Q2. But I'm still not sure why A2? My answer to my self is because A2 has 4 zeros and 4 ones which makes a new unique combinations. is my thought correct?
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SobiaholicNov 11 '12 at 22:55

'more correct' is a weird phrase for digital logic! If you write the full logc table you can see that You can see that A3=1 is the condition for the 3-to-8 decoder, and that its output depends on A2.
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Wouter van OoijenNov 12 '12 at 7:33

Thank you so much @WoutervanOoijen I get it now. I forgot that we're trying to run the 3to8 decoder. And to do that we used both Q3 and Q4 with a sum which makes it enabled. Thanks so much!!!!
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SobiaholicNov 13 '12 at 18:21

You can do this with two 3-8 decoders and get by with having to use just two parts of the same type. Common 3-8 decoders come with enables for both high and low polarities to make expansion a piece of cake. Excuse my quick hack drawing.